In Hilbert space $ \dim(\operatorname{coker}(T))=\dim(\ker(T^{*}))$

operator-theory

$H$ is a Hilbert space, and $T\in B(H)$ continuous and Fredholm operator.
Same books in definition of Fredholm use (when work in Banach space)

$\operatorname{ind}(T)=\dim(\ker(T))-\dim(\operatorname{coker}(T))$

and another's (when work with Hilbert space)

$\operatorname{ind}(T)=\dim(\ker(T))-\dim(\ker(T^{*}))$ , $T^{*}$ is the adjoint operator

so how can i show that

$\dim(\operatorname{coker}(T))=\dim(\ker(T^{*}))$

$\operatorname{coker}(T)=H/T(H)$ and $Ker(T^{*})=(T(H))^{\perp}$. so in Hilbert is true $H/T(H)\approx (T(H))^{\perp}$ ?

thanks

Best Answer

For any closed subspace $K$ of a Hilbert space $H$, we have $$H/K\cong K^\perp.$$

Proof.

$\newcommand{\co}{\operatorname{coker}}$Let \begin{align*}\Phi: & K^\perp\to H/K\\ & h\mapsto [h]\end{align*}

(1) Of course, $\Phi$ is linear bounded.

(2) $\Phi$ is injective. Suppose $h\in K^\perp$ such that $\Phi(h)=0$, then $h\in K^\perp\cap K$, so $h=0$.

(3) $\Phi$ is surjective. For any $h\in H$, there is some $h'\in K^\perp$ such that $h-h'\in K$, thus $\Phi(h')=[h']=[h]$.

Noting that $TH$ is closed subspace in the question, $H/TH\cong (TH)^\perp$.

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